rを|r|<1となる実体数とし、zn＝r^n(cos(nπ/4)+isin(nπ/4))とするときΣ

rを|r|<1となる実体数とし、zn＝r^n(cos(nπ/4)+isin(nπ/4))とするときΣ(n＝0〜∞)znの和を求めよ。さっぱり分かりません。教えてくださーい

No.6 ベストアンサー 回答者： mtrajcp 回答日時： 2022/04/19 11:16

Σ_{n=0～∞}zn

は
初項
1
公比
r(cos(π/4)+isin(π/4))
=r(1+i)/√2
の等比級数だから

Σ_{n=0～∞}zn
=
1/{1-r(1+i)/√2}
=
√2/{√2-r(1+i)}

No.5 回答者： mtrajcp 回答日時： 2022/04/19 04:06

Σ_{n=0～∞}zn

=
Σ_{k=0～∞}
{
r^(8k){cos(8kπ/4)+isin(8kπ/4)}
+r^(8k+1){cos((8k+1)π/4)+isin((8k+1)π/4)}
+r^(8k+2){cos((8k+2)π/4)+isin((8k+2)π/4)}
+r^(8k+3){cos((8k+3)π/4)+isin((8k+3)π/4)}
+r^(8k+4){cos((8k+4)π/4)+isin((8k+4)π/4)}
+r^(8k+5){cos((8k+5)π/4)+isin((8k+5)π/4)}
+r^(8k+6){cos((8k+6)π/4)+isin((8k+6)π/4)}
+r^(8k+7){cos((8k+7)π/4)+isin((8k+7)π/4)}
}
=
Σ_{k=0～∞}
{
r^(8k)
+r^(8k+1)(1+i)/√2
+r^(8k+2)i
+r^(8k+3)(-1+i)/√2
+r^(8k+4)(-1)
+r^(8k+5)(-1-i)/√2
+r^(8k+6)(-i)
+r^(8k+7)(1-i)/√2
}
=
{1+r(1+i)/√2+ir^2+r^3(-1+i)/√2-r^4+r^5(-1-i)/√2-ir^6+r^7(1-i)/√2}/(1-r^8)
=
{1-r^4}{1+r(1+i)/√2+ir^2+r^3(-1+i)/√2}/(1-r^8)
=
{1+r(1+i)/√2+ir^2+r^3(-1+i)/√2}/(1+r^4)

No.4 回答者： stomachman 回答日時： 2022/04/18 20:17

> rを|r|<1となる実体数と

「実体数」って何のこと？

> zn＝r^n(cos(nπ/4)+isin(nπ/4))

右辺は (r^n)(cos(nπ/4)+isin(nπ/4)) なのか r^(n(cos(nπ/4)+isin(nπ/4))) なのか。

No.3 回答者： mtrajcp 回答日時： 2022/04/18 20:04

Σ_{n=0～∞}zn

=
Σ_{k=0～∞}
{
r^(8k){cos(8kπ/4)+isin(8kπ/4)}
+r^(8k+1){cos((8k+1)π/4)+isin((8k+1)π/4)}
+r^(8k+2){cos((8k+2)π/4)+isin((8k+2)π/4)}
+r^(8k+3){cos((8k+3)π/4)+isin((8k+3)π/4)}
+r^(8k+4){cos((8k+4)π/4)+isin((8k+4)π/4)}
+r^(8k+5){cos((8k+5)π/4)+isin((8k+5)π/4)}
+r^(8k+6){cos((8k+6)π/4)+isin((8k+6)π/4)}
+r^(8k+7){cos((8k+7)π/4)+isin((8k+7)π/4)}
}
=
Σ_{k=0～∞}
{
r^(8k)
+r^(8k+1)(1+i)/√2
+r^(8k+2)i
+r^(8k+3)(-1+i)/√2
+r^(8k+4)(-1)
+r^(8k+5)(-1-i)/√2
+r^(8k+6)(-i)
+r^(8k+7)(1-i)/√2
}
=
{1+r(1+i)/√2+ir^2+r^3(-1+i)/√2-r^4+r^5(-1-i)/√2-ir^6+r^7(1-i)/√2}/(1-r^8)

No.2 回答者： masterkoto 回答日時： 2022/04/18 19:47

複素数平面を意識して考えてみては

No.1 回答者： ひなげしのはな 回答日時： 2022/04/18 17:32

私もですッ！